Question: Let $h$ be a function defined for all real numbers except for $-1$. Also let $h'$, the derivative of $h$, be defined as $h'(x)=\dfrac{(x+5)}{(x+1)}$. On which intervals is $h$ increasing? Choose 1 answer: Choose 1 answer: (Choice A) A $(-\infty,-5)$ and $(-1,\infty)$ (Choice B) B $(-5,-1)$ and $(-1,\infty)$ (Choice C) C $(-5,-1)$ only (Choice D) D $(-\infty,-5)$ only (Choice E) E The entire domain of $h$
Explanation: We can analyze the intervals where $h$ is increasing/decreasing by looking for the intervals where its derivative $h'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $h'(x)=\dfrac{(x+5)}{(x+1)}$ and that $h$ is undefined at $x=-1$. $h'(x)=0$ for $x=-5$. Our critical point is $x=-5$, and we should also consider $x=-1$. Our points divide the number line into three intervals: $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $(-\infty,\ \ \ \llap{-}5)$ $(\ \ \ \llap{-}5,\ \ \ \ \llap{-}1)$ $(\ \ \ \llap{-}1,\infty)$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $(-\infty,-5)$ $x=-6$ $h'(-6)=\dfrac{1}{5}>0$ $h$ is increasing $\nearrow$ $(-5,-1)$ $x=-2$ $h'(-2)=-3<0$ $h$ is decreasing $\searrow$ $(-1,\infty)$ $x=0$ $h'(0)=5>0$ $h$ is increasing $\nearrow$ In conclusion, $h$ is increasing over the intervals $(-\infty,-5)$ and $(-1,\infty)$.